# How to find the limit for lim x -> 0 ( 1/(xsqrtx) * int_0^sqrtx cos(pi/2*e^(t^2))dt) with l'Hospital?

Jun 11, 2018

$- \frac{\pi}{6}$

#### Explanation:

${\lim}_{x \to 0} \frac{{\int}_{0}^{\sqrt{x}} \cos \left(\frac{\pi}{2} \cdot {e}^{{t}^{2}}\right) \mathrm{dt}}{x \sqrt{x}}$

That is $\frac{0}{0}$ indeterminate so we use LHR and differentiate top and bottom wrt x.

For the top use the Liebnitz Integral Rule, which applies here as:

• $\boldsymbol{\frac{d}{\mathrm{dx}} \left({\int}_{0}^{v \left(x\right)} f \left(s\right) \setminus \mathrm{ds}\right) = f \left(v \left(x\right)\right) \cdot {v}^{'} \left(x\right)}$

$= {\lim}_{x \to 0} \frac{\cos \left(\frac{\pi}{2} \cdot {e}^{x}\right) \cdot \frac{1}{2 \sqrt{x}}}{\frac{3}{2} \sqrt{x}}$

$= {\lim}_{x \to 0} \frac{\cos \left(\frac{\pi}{2} \cdot {e}^{x}\right)}{3 x}$

That is still $\frac{0}{0}$ indeterminate so we go again with LHR.

$= {\lim}_{x \to 0} \frac{\frac{\pi}{2} {e}^{x} \cdot - \sin \left(\frac{\pi}{2} \cdot {e}^{x}\right)}{3} = \frac{\left(\frac{\pi}{2}\right) \left(- \sin \left(\frac{\pi}{2}\right)\right)}{3} = - \frac{\pi}{6}$